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I'm new to learning about electricity. I have learned that Electrons moves from - to + when there is a voltage differential.

My conclusion is, that I can connect the plus side of one battery to a bulb, and connect it to the minus side of the second battery without closing a circuit.

This way, the electrons will move, and the bulb will light?

Is it will light for at least a few nanoseconds?

enter image description here

Bonus: There is an online website that can help me create the diagram and test it? or just share here the image?

Is it the truth that the bulb will be lighted for a few nanoseconds? (see answers below)

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  • \$\begingroup\$ First part: Yes, only in electronics we don't care about the actual direction. Second part: the circuit discribed is a closed circuit. Simulation: electronics.stackexchange.com/questions/154252/… \$\endgroup\$ – JWRM22 Mar 2 '17 at 10:19
  • \$\begingroup\$ @JWRM22 This isn't a closed circuit. Using the positive of one battery with the negative of a second battery? Whats the negative of the first battery connected to in that case? \$\endgroup\$ – Doodle Mar 2 '17 at 10:24
  • \$\begingroup\$ You have merely connected the two batteries in series via an unnecessarily thin wire. \$\endgroup\$ – Brian Drummond Mar 2 '17 at 11:03
  • \$\begingroup\$ @Aminadav: why do you think it should be possible to light the bulb without closing a circuit? Why do you think electrons would move? \$\endgroup\$ – Curd Mar 2 '17 at 13:45
  • \$\begingroup\$ @Hayman, you are correct. I've read the question wrong, also a good reason to add a schematic. \$\endgroup\$ – JWRM22 Mar 3 '17 at 14:27
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You may have a look at this site: http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/electrochem.html enter image description here

"In order for the voltaic cell to continue to produce an external electric current, there must be a movement of the sulfate ions in solution from the right to the left to balance the electron flow in the external circuit"

Because that ions can't move between the two batteries, your bulb will not produce light.

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  • \$\begingroup\$ "Must be a movement of the sulfate ions in solution from the right to the left to balance the electron flow in the external circuit" Why? \$\endgroup\$ – Aminadav Glickshtein Mar 2 '17 at 10:28
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    \$\begingroup\$ If you don't have a complete, circular path for current flow, none flows. End of story. That is the "WHY". It's like laying on the ground. You can't fall, you are already at the lowest possible energy potential. Now if someone pulled the ground out from under you, you would fall to the new ground level, releasing energy. \$\endgroup\$ – R Drast Mar 2 '17 at 11:28
  • \$\begingroup\$ The reason is that the ability of the battery to displace charge depends on a dynamical equilibrium between electrode and solution. When you 'remove' one electron from the zinc electrode, you alter this delicate equilibrium and in order to re-estabilish it you must remove (or, if you wish, neutralize) a Zn++ ion. The SO4-- ion coming from the other side will do that more than gladly. \$\endgroup\$ – Sredni Vashtar Mar 2 '17 at 13:17
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Battery terminals are marked (+) and (-) because the RELATIVE voltages are in that relation. Until you connect the two batteries together at some terminal, you don't know the absolute voltage of battery #1's (-) terminal is more negative than battery #2's (+) terminal.

While one terminal of a battery is floating (unconnected) there's no way to apply the Kirchoff rule that all voltages in a closed circuit loop add to zero, so there's no way to identify a resistor (or lamp filament) as having an applied voltage.

In a related note, if you attach the two (open) battery terminals with a very-high-value resistor (a million ohms?), the lamp will get some current, but not enough to heat it and make a glow. An interesting question is, what resistor value WOULD make the light glow.

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The misunderstanding is thinking that there is an imbalance of electrons. There isn't on batteries. Both the positive and negative terminals are entirely normal metal with equal numbers of positive and negative electrons present.

The difference is in the electrochemical potential between the electrolyte and the electrodes. This creates an electric field with a potential difference between the two electrodes. This is not something with an absolute reference, it's entirely relative.

(The only situation where you do get a significant mismatch in electron count is electrostatically charged items, like Ben Franklin's rods rubbed with wool. In this case you can get a current to flow for a very short time between two charged items without having a complete circuit.)

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  • \$\begingroup\$ I'm not so sure about the "metal with equal number of positive [ions] and negative electrons present". While you are right in stating the the battery as a whole is electrically neutral and the positive and negative poles do not accumulate charge of opposite polarity, the electrodes experience an imbalance in charge. The point is that this imbalance in counterbalanced by the ions in the electrolyte and therefore one would not see accumulation of charge at the poles. That's why batteries do not implode :-) due to the strenght of the electrostatic force. \$\endgroup\$ – Sredni Vashtar Mar 2 '17 at 13:59
  • \$\begingroup\$ Exactly. Thanks. I have learned about electrostatic. Do you have a good link for learn more about potential difference? \$\endgroup\$ – Aminadav Glickshtein Mar 2 '17 at 14:04
  • \$\begingroup\$ Old technology links only. They were called "books". Kip "Fundamentals of Electricity and magnetism", 2nd edition does a reasonable job of explaining in a couple of pages how a battery works from a physicist point of view. And then there are chemistry books - I cannot find one much better than the other though, so anything goes (Whitten, Atkins-Jones, Mahan-Myers...) \$\endgroup\$ – Sredni Vashtar Mar 2 '17 at 14:08
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In order to draw the current the circuit must be closed meaning there should be easier path of current to flow in a circuit. In your case as there is no closed path there will be no current (flow of electrical charge) flowing, thus the bulb will not glow

Also in case of two different batteries the reference is different, unless if they both have common reference (e.g same ground) that means they are connected either in parallel or series which in terms make the current flow and bulb will glow.

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Hopefully this is something that helps you

enter image description here

The current needs a closed conductive path. In the vacuum the electrons can propagate without a conductor, but not in the air.

Let's imagine that the middle scenario makes light (=current) Then the total amount of electrons would grow at the other battery. Finally they will push so strongly the new ones back that the current stops.

ADDENDUM: But how long the current exists and how much electrons flow through the bulb if we put together the system in the middle? For that exists the concept named"capacitance". Imagine to put the system together instantneously (=in no time) Then the electric fields of the batteries push and pull the electrons until new balance is found. There really is a current until the capacitance between X and Y is fully charged. Assuming the capacitance to be a few picofarads and the total resistance in the parts max. a couple of Ohms, the current stops well before one nanosecond. The total moved charge is well below one nanocoulomb. If the battery voltage is 12 volts, then the total dissipation in the bulb is so low that no observable filament warmup exists in the bulb - no light, but a short current pulse yes, but in practice also not observable.

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  • \$\begingroup\$ Thanks! The question is WHY electrons do not pass from the + to the - in the middle image. The are a lots of electrons in the left battery and not enough electrons in the right battery. They want to be in a balance, so they should move! \$\endgroup\$ – Aminadav Glickshtein Mar 2 '17 at 10:36
  • \$\begingroup\$ So, how much time do you have until the current stops? You will get a 1 second or 1ms of light? \$\endgroup\$ – Aminadav Glickshtein Mar 2 '17 at 10:44
  • \$\begingroup\$ There is no mismatch in the count of electrons on batteries. \$\endgroup\$ – pjc50 Mar 2 '17 at 12:15
  • \$\begingroup\$ @Aminadav Moved my chatting to the answer and completed it, too. \$\endgroup\$ – user287001 Mar 2 '17 at 13:37
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You have made a good start in recognizing that electrons that are moving cause the lamp to light. Your circuit is in static equilibrium (redrawn below with stackexchange's menu-bar circuit-drawing facility):

schematic

simulate this circuit – Schematic created using CircuitLab
No electrons move in the above circuit. The negative end of BAT1 is at the same voltage as the positive end of BAT2. So no electrons move through the lamp, and there's no light.
However, if you measure the voltage from the positive-end of BAT1, to the negative end of BAT2, you'll measure 6+9=15 volts. Both batteries need to do no work.

Perhaps your error is in assuming that these loose ends are at the same potential voltage. If you force them to be at the same voltage by connecting them with a low-resistance wire, electrons now flow (revised circuit below). LAMP1 has 15 volts across its terminals. This is a series circuit, with BAT1, LAMP1, and BAT2 experiencing the same electron flow (current).

schematic

simulate this circuit

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